Integrating absolute values (definite integrals) - can't find good info

**TYTY** · January 31st 2009, 10:44 AM

Anyone have a good tutorial? Google just sends me to people asking questions like I am. I know it's supposed to be split up. Anyways, here is the problem I am stuck on:

from -2 to 2 integrate (1 - 3|x|)dx

Whenever I think about splitting it up on a definite integral, my brain sort of melts. My book is no help either

**galactus** · January 31st 2009, 10:52 AM

Yes, when dealing with absolute values, break them up.

All it is is two lines. One for negative and one for positive.

$1-3|x|$ is 1+3x and 1-3x

$\int_{-2}^{0}(1+3x)dx+\int_{0}^{2}(1-3x)dx$

**TYTY** · January 31st 2009, 10:56 AM

Originally Posted by galactus

Yes, when dealing with absolute values, break them up.

All it is is two lines. One for negative and one for positive.

$1-3|x|$ is 1+3x and 1-3x

$\int_{-2}^{0}(1+3x)dx+\int_{0}^{2}(1-3x)dx$

clear, concise

thank you

**Soroban** · January 31st 2009, 11:08 AM

Hello, TYTY!

With absolute values, a sketch usually helps . . .

$\int^2_{-2}\left(1-3|x|\right)\,dx$

We know the graph of $y = |x|$ . . . a V with vertex at the origin.

The graph of $y = 3|x|$ . . . "climbs steeper".

The graph of $y \:=\:-3|x|$ . . . inverts the previous graph.

The graph of $y \:=\:1-3|x|$ . . . moved up one unit.

Code:

              |
              *
             /|\
            /:|:\
     -2    /::|::\    2
  - - + - o - + - o - + - -
      |::/    |    \::|
      |:/     |     \:|
      |/      |      \|
      *       |       *
              |

We have an inverted V, vertex at (0,1), x-intercepts at $\pm\tfrac{1}{3}$

They ask for the "area" from $x = \text{-}2$ to $x=2,$
. . and some of it is "negative area".
We don't need Calculus to find the areas of triangles.

[My answer is: -8]

**TYTY** · January 31st 2009, 11:40 AM

Originally Posted by Soroban

Hello, TYTY!

With absolute values, a sketch usually helps . . .

We know the graph of $y = |x|$ . . . a V with vertex at the origin.

The graph of $y = 3|x|$ . . . "climbs steeper".

The graph of $y \:=\:-3|x|$ . . . inverts the previous graph.

The graph of $y \:=\:1-3|x|$ . . . moved up one unit.

Code:

              |
              *
             /|\
            /:|:\
     -2    /::|::\    2
  - - + - o - + - o - + - -
      |::/    |    \::|
      |:/     |     \:|
      |/      |      \|
      *       |       *
              |

We have an inverted V, vertex at (0,1), x-intercepts at $\pm\tfrac{1}{3}$

They ask for the "area" from $x = \text{-}2$ to $x=2,$
. . and some of it is "negative area".
We don't need Calculus to find the areas of triangles.

[My answer is: -8]

This is a good point and also how the text wanted it to be done. When I thought about integrating it is where I tripped up though.

Thread: Integrating absolute values (definite integrals) - can't find good info

LinkBack

Thread Tools

Display

Integrating absolute values (definite integrals) - can't find good info

Similar Math Help Forum Discussions

Derivatives, Integrals, Absolute Values Oh My!

Find the absolute maximum and absolute minimum values of f on the given interval.

Find the absolute maximum and absolute minimum values of the function

Need help with integrals with absolute values

Integrating Absolute values

Search Tags